`n_{Br_2}=0,15.2=0,3(mol)`

`PTHH:C_2H_4+Br_2->C_2H_4Br_2`

`(mol)`   `0,3`        `←0,3→`              `0,3`

`a.V_{C_2H_4}=0,3.22,4=6,72(l)`

`%V_{C_2H_4}={6,72}/{11,1555}.100%~~60,24%`

`%V_{CH_4}=100-60,24=39,76%`

`n_{CH_4}={11,1555-6,72}/{22,4}~~0,2(mol)`

`b.PTHH:CH_4+2O_2 \overset(t^o)(->)CO_2+2H_2O`

`(mol)`     `0,2->`      `0,4`

             `C_2H_4+3O_2 \overset(t^o)(->)2CO_2+2H_2O`

`(mol)`      `0,3->`             `0,9`

`V_{O_2}=(0,4+0,9).22,4=29,12(l)`

`a)`

Sửa `11,1555g->11,1555L`

`C_2H_4+Br_2->C_2H_4Br_2`

`->n_{C_2H_4}=n_{Br_2}=0,15.2=0,3(mol)`

`\%V_{C_2H_4}={0,3.24,79}/{11,1555}.100\%=66,67\%`

`\%V_{CH_4}=100-66,67=33,33\%`

`b)`

`n_{CH_4}={11,1555-0,3.24,79}/{24,79}=0,15(mol)`

`CH_4+2O_2` $\xrightarrow{t^o}$ `CO_2+2H_2O`

`C_2H_4+3O_2` $\xrightarrow{t^o}$ `2CO_2+2H_2O`

`->n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=1,2(mol)`

`V_{O_2}=1,2.24,79=29,748(L)`